1. **State the problem:** We are given vectors $v_1 = (1,2,1)$, $v_2 = (2,9,0)$, and $v_3 = (3,3,4)$ which form a basis $S$ for $\mathbb{R}^3$. We want to find the coordinate vector of $v = (5,-1,9)$ relative to the basis $S$. 2. **Formula and explanation:** The coordinate vector $[v]_S = (a,b,c)$ satisfies $$a v_1 + b v_2 + c v_3 = v$$ which means $$a(1,2,1) + b(2,9,0) + c(3,3,4) = (5,-1,9).$$ We need to solve for $a,b,c$. 3. **Set up the system of equations:** $$\begin{cases} a + 2b + 3c = 5 \\ 2a + 9b + 3c = -1 \\ a + 0b + 4c = 9 \end{cases}$$ 4. **Solve the system:** From the third equation: $$a + 4c = 9 \implies a = 9 - 4c.$$ Substitute $a$ into the first and second equations: - First: $(9 - 4c) + 2b + 3c = 5 \implies 9 + 2b - c = 5 \implies 2b - c = -4.$ - Second: $2(9 - 4c) + 9b + 3c = -1 \implies 18 - 8c + 9b + 3c = -1 \implies 9b - 5c = -19.$ 5. **Solve for $b$ and $c$:** From $2b - c = -4$, express $c$: $$c = 2b + 4.$$ Substitute into $9b - 5c = -19$: $$9b - 5(2b + 4) = -19 \implies 9b - 10b - 20 = -19 \implies -b - 20 = -19 \implies -b = 1 \implies b = -1.$$ Then, $$c = 2(-1) + 4 = -2 + 4 = 2.$$ 6. **Find $a$:** $$a = 9 - 4c = 9 - 4(2) = 9 - 8 = 1.$$ 7. **Final answer:** The coordinate vector of $v$ relative to basis $S$ is $$[v]_S = (a,b,c) = (1, -1, 2).$$